Difference between revisions of "2017 AIME II Problems/Problem 8"
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− | We start with the last two terms of the polynomial <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</math>, which are <math>\frac{n^5}{5!}+\frac{n^6}{6!}</math>. This can simplify to <math>\frac{6n^5+n^6}{720}</math>, which can further simplify to <math>\frac{n^5(6+n)}{720}</math>. Notice that the prime factorization of <math>720</math> is <math>5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2</math>. In order for <math>\frac{n^5(6+n)}{720}</math> to be an integer, one of the parts must divide <math>5, 3</math>, and <math>2</math>. Thus, one of the parts must be a multiple of <math>5, 3</math>, and <math>2</math>, and the LCM | + | We start with the last two terms of the polynomial <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</math>, which are <math>\frac{n^5}{5!}+\frac{n^6}{6!}</math>. This can simplify to <math>\frac{6n^5+n^6}{720}</math>, which can further simplify to <math>\frac{n^5(6+n)}{720}</math>. Notice that the prime factorization of <math>720</math> is <math>5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2</math>. In order for <math>\frac{n^5(6+n)}{720}</math> to be an integer, one of the parts must divide <math>5, 3</math>, and <math>2</math>. Thus, one of the parts must be a multiple of <math>5, 3</math>, and <math>2</math>, and the LCM of these three numbers is <math>30</math>. This means <cmath>n^5 \equiv 0\pmod{30}</cmath> or <cmath>6+n \equiv 0\pmod{30}</cmath> Thus, we can see that <math>n</math> must equal <math>0\pmod{30}</math> or <math>0\pmod{30}-6</math>. Note that as long as we satisfy <math>\frac{6n^5+n^6}{720}</math>, <math>2!, 3!</math>, and <math>4!</math> will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. <math>4! = 2\cdot2\cdot2\cdot2\cdot3</math>, and this will be divisible by <math>2^4\cdot3^4\cdot5^4</math>. Now, since we know that <math>n</math> must equal <math>0\pmod{30}</math> or <math>0\pmod{30}-6</math> in order for the polynomial to be an integer, <math>n\equiv0, 24\pmod{30}</math>. To find how many integers fulfill the equation and are <math><2017</math>, we take <math>\lfloor\frac{2017}{30}\rfloor</math> and multiply it by <math>2</math>. Thus, we get <math>67\cdot2=\boxed{134}</math>. |
Solution by IronicNinja~ | Solution by IronicNinja~ |
Revision as of 13:21, 20 May 2020
Contents
[hide]Problem
Find the number of positive integers less than such that is an integer.
Solution 1
We start with the last two terms of the polynomial , which are . This can simplify to , which can further simplify to . Notice that the prime factorization of is . In order for to be an integer, one of the parts must divide , and . Thus, one of the parts must be a multiple of , and , and the LCM of these three numbers is . This means or Thus, we can see that must equal or . Note that as long as we satisfy , , and will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. , and this will be divisible by . Now, since we know that must equal or in order for the polynomial to be an integer, . To find how many integers fulfill the equation and are , we take and multiply it by . Thus, we get .
Solution by IronicNinja~
Solution 2 (Not Rigorous)
Writing the last two terms with a common denominator, we have By inspection. this yields that . Therefore, we get the final answer of .
Solution 3
Taking out the part of the expression and writing the remaining terms under a common denominator, we get . Therefore the expression must equal for some positive integer . Taking both sides mod , the result is . Therefore must be even. If is even, that means can be written in the form where is a positive integer. Replacing with in the expression, is divisible by because each coefficient is divisible by . Therefore, if is even, is divisible by .
Taking the equation mod , the result is . Therefore must be a multiple of . If is a multiple of three, that means can be written in the form where is a positive integer. Replacing with in the expression, is divisible by because each coefficient is divisible by . Therefore, if is a multiple of , is divisibly by .
Taking the equation mod , the result is . The only values of that satisfy the equation are and . Therefore if is or mod , will be a multiple of .
The only way to get the expression to be divisible by is to have , , and . By the Chinese Remainder Theorem or simple guessing and checking, we see . Because no numbers between and are equivalent to or mod , the answer is .
Solution 4
Note that will have a denominator that divides . Therefore, for the expression to be an integer, must have a denominator that divides . Thus, , and . Let . Substituting gives . Note that the first terms are integers, so it suffices for to be an integer. This simplifies to . It follows that . Therefore, is either or modulo . However, we seek the number of , and . By CRT, is either or modulo , and the answer is .
-TheUltimate123
Step Solution
Clearly is an integer. The part we need to verify as an integer is, upon common denominator, . Clearly, the numerator must be even for the fraction to be an integer. Therefore, is even and n is even, aka for some integer . Then, we can substitute and see that is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get . It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the , and we see that for some integer . From there we now know that . If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that , so combining with divisibility by 6, is or (mod ). There are cases for each, hence the answer .
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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